by using [1]},
we obtain by (REF ),
\(\begin{split}&\int _0^{\infty }\left(\int _{-1}^1(r^2+2rs+1)^q\log (r^2+2rs+1)(1-s^2)^{(d-3)/2}\,{\rm d}s\right)\,\frac{{\rm d}r}{r^{\alpha +1}}\\&=\frac{1}{2}\sum _{n=0}^{\infty }\binom{q}{2n}2^{2n}B\left(n+\frac{1}{2},\frac{d-1}{2}\right)B\left(n-\frac{\alpha }{2},n+\frac{\alpha }{2}-q\right)\left[\psi (2n-q)-\psi \left(n+\frac{\alpha }{2}-q\right)\right]\\&\qquad -\frac{1}{2}\sum _{n=1}^{\infty }\left[\sum _{k=0}^{2n-1}\binom{q}{k}\frac{(-1)^k}{2n-k}\right]2^{2n}B\left(n+\frac{1}{2},\frac{d-1}{2}\right)B\left(n-\frac{\alpha }{2},n+\frac{\alpha }{2}-q\right).\end{split}\) 
